BASIC ELECTRONICS DC CIRCUIT ANALYSIS. December 2011

Transcription

1 AM BASIC ELECTRONICS DC CIRCUIT ANALYSIS December 2011 DISTRIBUTION RESTRICTION: Approved for public release. Distribution is unlimited. DEPARTMENT OF THE ARMY MILITARY AUXILIARY RADIO SYSTEM FORT HUACHUCA ARIZONA

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3 AM Basic Electronics - DC Circuit Analysis CHANGE PAGE LIST OF EFFECTIVE PAGES INSERT LATEST CHANGED PAGES. DISTROY SUPERSEDED PAGES NOTE The portion of this text affected by the changes is indicated by a vertical line in the outer margins of the page. Changes to illustrations are indicated by shaded or screened areas or by miniature pointing hands. Changes of issue for original and changed pages are: ORIGIONAL..0. Page Chang NO. e No. Title Page NO. Change No. Page No. Change No. *Zero in this column indicates an original page A Change 0 US Army 2. RETAIN THIS NOTICE AND INSERT BEFORE TABLE OF CONTENTS. 3. Holders of this document will verify that page changes and additions indicated above have been entered. This notice page will be retained as a check sheet. This issuance, together with appended pages, is a separate publication. Each notice is to be retained by the stocking points until the standard is completely revised of canceled. Ver. 1.0 iii

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5 AM Basic Electronics - DC Circuit Analysis CONTENTS 1 DC CIRCUIT ANALYSIS REFERENCE OHM'S LAW TEMPERATURE COEFFICIENT HEAT AND POWER POWER DISSIPATION IN RESISTORS THE SERIES CIRCUIT PARALLEL CIRCUITS VOLTAGE DIVIDERS COMBINATION SERIES/PARALLEL CIRCUITS Ver. 1.0 v

6 AM Basic Electronics DC Circuits Analysis IMPROVEMENTS (Suggested corrections, or changes to this document, should be submitted through your State Director to the Regional Director. Any Changes will be made by the National documentation team. DISTRIBUTION Distribution is unlimited. VERSIONS The Versions are designated in the footer of each page if no version number is designated the version is considered to be 1.0 or the original issue. Documents may have pages with different versions designated; if so verify the versions on the Change Page at the beginning of each document. REFERENCES The following references apply to this manual: Allied Communications Publications (ACP): ACP Glossary of Communications Electronics Terms US Army FM/TM Manuals 1. TM Electrical Design, Lightning and Static Electricity Protection 2. TM Facilities Engineering Electrical Facilities Safety 3. TM Grounding and Bonding in Command, Control, Communications, Computer, Intelligence, Surveillance, and Reconnaissance (C4ISR) Facilities 4. TM Electrical Fundamentals, Direct Current 5. TM-664 Basic Theory and Use of Electronic Test Equipment US Army Handbooks 1. MIL-HDBK Grounding, Bonding and Shielding Design Practices Commercial References 2. Basic Electronics, Components, Devices and Circuits; ISBN X, By William P Hand and Gerald Williams Glencoe/McGraw Hill Publishing Co. 3. Standard Handbook for Electrical Engineers - McGraw Hill Publishing Co. CONTRIBUTORS This document has been produced by the Army MARS Technical Writing Team under the authority of Army MARS HQ, Ft Huachuca, AZ. The following individuals are subject matter experts who made significant contributions to this document. William P Hand vi Ver. 1.0

7 AM Basic Electronics - DC Circuit Analysis 1 DC CIRCUIT ANALYSIS 1.1 REFERENCE Basic Electronics, Components, Devices and Circuits; ISBN X By William P Hand and Gerald Williams. 1.2 OHM'S LAW The German physicist George Simon Ohm conducted extensive experiments into the abilities of various materials to conduct electric energy. He found that materials with the same physical size had different resistances. He was able to establish that, for a fixed voltage, the amount of current flowing through any material depends upon type and physical size of material. The ratio of voltage and current (resistance) for any given type of conductor is a constant that is dependent upon material and its dimensions. This ratio can be expressed mathematically as follows: E= IR In this equation, E is the applied voltage in volts, I is the current in amperes, and R is the resistance in ohms, the unit of measure for resistance. The equation can be interpreted as: 1 volt will force 1 ampere of current through a resistance of 1 ohm, When it is desired to specify how good a conductor is, rather than how much resistance it offers, the unit is the mho (ohm spelled backward), Electric conductance/siemens The siemens is the practical unit of conductance, The siemens is the reciprocal of ohm since conductance is the reciprocal of resistance, The relationship between ohms and siemens is given by G = R siemens, A conductance of one siemens will permit a current flow of one ampere (A) under an electrical pressure of one volt (V), the siemens is a new unit honoring a pioneer in electricity, NOTE Formerly, the unit of conductance was called the mho. The symbol for conductance is G, and the abbreviation for siemens is S, The symbol for ohm is the Greek letter Ω (omega). The symbol for the mho is an upside-down omega, U. The three forms of Ohm's law are as follows (Reference Fig. 1-1): Ver

8 AM Basic Electronics DC Circuits Analysis Figure 1-1 The Ohm's Law Memory Aid The voltage (E)* that is required to maintain a certain current in a circuit in which resistance is known is equal to the product of the current (I) and the resistance (R). E = IR The current (I) in any circuit is equal to the voltage (E) applied, divided by resistance of the circuit. The resistance (R) needed in a circuit so that a certain current (I) will flow when a certain voltage (E) is applied is equal to the voltage (E) divided by the current (I). * E stands for electromotive force. V (for voltage) is also commonly used (V = I R, I = V / R, R = V / I). In order to understand applications of the Ohm's law variations, we will work an example of each type. (1) Example 3-1 Problem: What voltage must be applied to a circuit containing a resistance of 150 ohms and a required current of 3 amps in the circuit? 1-2 Ver. 1.0

9 AM Basic Electronics - DC Circuit Analysis Solution: E = IR = (3) (150) = 450 volts (2) Example 3-2 Problem: If a lamp with a resistance of 75 ohms is connected across a voltage source of 117 volts, what current will flow through the lamp? Solution: Ver

10 AM Basic Electronics DC Circuits Analysis (3) Example 3-3 Problem: What resistance will cause a current of 1.2 amps to flow when a voltage of 150 volts is applied to the circuit? Solution: Up to this point we have not considered temperature in our calculations. We have considered only a normal room temperature of 20 degrees Celsius (20 C). The resistance of any conductor depends upon the number of free electrons in it. Therefore, as the temperature of a conductor is raised, energy is transferred to the atoms because of the temperature increase. This additional energy given to the atoms causes them to move around more than they did, thus causing an increase in distance between atoms. The increase in distance will provide more space for free electrons to move in without colliding with an atom. 1-4 Ver. 1.0

11 AM Basic Electronics - DC Circuit Analysis 1.3 TEMPERATURE COEFFICIENT When we heat most substances they will enlarge to some extent. As we increase emperature, it will cause an increase in resistance in most substances. This is known as a positive temperature coefficient. Reference the graph shown in Figure 1-2 for copper. Note that it is approximately linear (a straight line graph) over the normal temperature range in which electronic equipment operates. Figure 1-2 Temperature Coefficient of Copper (Cu) Scientists have measured the resistance of conductors at varying temperatures and have determined that most metals increase in resistance. The increase is approximately linear over a wide temperature range. Ver

12 AM Basic Electronics DC Circuits Analysis NOTES 1-6 Ver. 1.0

13 AM Basic Electronics - DC Circuit Analysis 2 HEAT AND POWER When current flows in a circuit there is some power (or J2 R) loss in the circuit. This loss is in the form of heat. (Power) Watts = I 2 R Commercial resistors have two basic ratings, the resistance in ohms and a power rating expressed in watts. A resistor must be able to dissipate the heat it produces. If the internal temperature of a resistor increases to a high level its resistance will change, and often it is a permanent change. Most common lower wattage resistors, that is, ¼ to 2 watts, have their resistance values specified by a color code printed on their bodies. The wattage rating is simply a matter of the resistor's physical size, which, with a little practice, can easily be learned. Table 2-1 shows the resistor color code and how it works. The color code should be memorized, as it will be used in almost all electronic equipment you will encounter. 1. Example 4-1 Problem: Figure the value of a resistor with the following color bands: red, first band; green, second band; orange, third band; silver, fourth band. Reference Resistor Color Code Chart in Table 1-1 Red Green Orange Silver (3 zeros) 10% Solution: 25,000 ohms ± 10% 2.1 POWER DISSIPATION IN RESISTORS The power dissipated by a resistor is the product of current through it and voltage across it. P= E I This formula can also be stated in two other forms: P=I 2 R 1. Example 5-1 Problem: Find power dissipation of a resistor with an applied voltage of 10 volts and a current of 0.2 amps (the unit of measure is the watt). Solution: P = 10 x 0.2 = 2 watts Ver

14 AM Basic Electronics DC Circuits Analysis Table 2-1 Resistor Color Code 2-2 Ver. 1.0

15 AM Basic Electronics - DC Circuit Analysis The power dissipation can also be calculated when only current and resistance or voltage and resistance are known. 2. Example 5-2 Problem: Find power dissipation of a 100.Ω resistor with a current through it of 0.1 amp. Solution: P=I 2 R P=0.1 2 x 100 P= 1 watt 3. Example 5-3 Problem: Find the power dissipation of a 100 ohm resistor when there is a voltage of 10 volts across it. Solution: Ver

16 AM Basic Electronics DC Circuits Analysis NOTES 2-4 Ver. 1.0

17 AM Basic Electronics - DC Circuit Analysis 3 THE SERIES CIRCUIT In a series circuit the same current passes through EACH element in the circuit before completing its path to the source. A simple series circuit is shown in Figure 3-1 (A). This circuit is made up of six lamps in series, like a string of Christmas tree lamps. In order for the current to complete its path, it must flow through each lamp in turn before flowing back to the battery. If anyone of the six lamps burns out, it will open the circuit and current will cease the flow. This opening of the circuit is like a switch. A switch is a simple device to open a circuit whenever current flow is not wanted. In part A of the figure, the circuit is redrawn showing resistors in place of the lamps. Many times a circuit such as that shown in part B of the figure can be more easily understood when symbols of an equivalent nature are used as shown in Figure 3-2. Figure 3-1 Series DC Circuit Ver

18 AM Basic Electronics DC Circuits Analysis Figure 3-2 Equivalent Series Circuit 3-2 Ver. 1.0

19 AM Basic Electronics - DC Circuit Analysis 4 PARALLEL CIRCUITS In Figure 4-1 we can see the differences in series and parallel circuits. In part A of the figure there is only one path for current to follow. This single path through two resistors and back to the battery makes it a series circuit. In part B of the figure, we have a much different situation. There are two different paths that current can, and does, or follow. Currents II and h both flow from the battery to the junction of the two resistors. At that point the current splits; current II will flow through RI and current will flow through R2. The two currents then recombine after flowing through the resistors and flow back to the battery, thus completing the circuit. A very important distinction should be noted here about circuits. It is not totally correct to state that two components connected across each other make a parallel circuit. As you can see in Figure 6-1, both series and parallel circuits across the power source, in this case, the battery. In part A, the series circuit, as we already know, the same current is flowing through both resistors and the battery. In part B of the figure, RI and R2 are connected across each other and the combination of the two resistors is connected across the battery. The two resistors RI and R2 cannot be in series because different currents flow through each resistor. It can be seen that the way in which the current flows in a circuit determines whether it is a series or a parallel circuit; and that the total current or combined current, is the sum of the currents flowing through each branch, as shown by I t = I 1 + I 2 + i In parallel circuits. the total current flowing through any parallel combination can always be found by applying Ohm's law to each branch in turn and then adding the resulting currents. Figure 4-1 Series and Parallel Circuits Example 4-1 Problem: What is the total current flowing, when a 200 ohm, a 150 ohm, and Ver

20 AM Basic Electronics DC Circuits Analysis a 300 ohm resistor are connected in parallel across a 100 volt source? Solution: 100 II = 200 = = 150 = = 300 = 0.33 It = = 1.49 amps From this example it can be seen that the total current is much greater than the current through any one of the individual parallel branches. We now have enough data to state Kirchhoff s first law: At any junction point in an electrical circuit, the algebraic sum of the currents entering the point must be equal to the algebraic sum of the currents leaving the point. Let us look at an example of this law. Example 4-2 Problem: What is the current through the resistor R 2 in the circuit? Diagram for example 4-2. Solution: Therefore, from Kirchhoffs Law 4-2 Ver. 1.0

21 AM Basic Electronics - DC Circuit Analysis In the examples it was shown that the branches is the lighting circuit in a house. Each light in the house is connected in parallel with all the others to the 117 volt source, and switching one of the lights on or off has no effect on the others in the house. Ver

22 AM Basic Electronics DC Circuits Analysis NOTES 4-4 Ver. 1.0

23 AM Basic Electronics - DC Circuit Analysis 5 VOLTAGE DIVIDERS The basic voltage divider normally consists of a series of resistors having two input terminals, across which the input voltage from the power source is applied. The outputs are across the various resistors depending upon what part of the input voltage is desired. The simple voltage divider is a special series circuit. Shown in Figure 5-1 is an example of a voltage divider. Example 5-1 Problem: Figure 5-1 shows a simple voltage divider. Find the voltage between each terminal and ground. Solution: a. Calculate total current I t = E/(R l + R 2 + R 3 ) (Ohm's law) I t = 100/1000 = 0.1 amp Using the resistance between terminal A and ground and the total current, calculate the voltage E= IR E= 0.1 X 100 E = 10 volts Using the total resistance between terminal B and ground and the total current, calculate the voltage E= IR E= 0.1 X E= 0.1 X 400 Figure 5-1 Simple Voltage Divider. E = 40 volts Ver

24 AM Basic Electronics DC Circuits Analysis There is no need to calculate the voltage between terminal C and ground because it is directly across the battery. Suppose we increase the value of resistors R I, R 2, and R 3 to 6k, 3k, and lk, respectively, and solve the problem again. a. Calculate the total current. Keep in mind that k means 1,000. b. Using the resistance between terminal A and ground and the total current, calculate the voltage. E= IR E= 0.01 X 1000 E = 10 volts c. Using the total resistance between terminal B and ground and the total current, calculate the voltage. E= IR E = 0.01 X E = 0.01 X 4000 E = 40 volts Voltage between terminal and first second ground example example A 10 volts 10 volts B 40 volts 40 volts C 100 volts 100 volts The two examples make the important point that the terminal voltages in a voltage divider depend only upon the resistance ratios and not on the absolute values of the resistors. 5-2 Ver. 1.0

25 AM Basic Electronics - DC Circuit Analysis 6 COMBINATION SERIES/PARALLEL CIRCUITS As the name implies, the series-parallel circuit is a combination of series and parallel circuits. Figure 6-1 shows the two basic types of series-parallel circuits and their simplified equivalent circuits. As can be seen from Figure 6-1, a series-parallel circuit can be defined as one that has the combined characteristics of both series circuits and parallel circuits in the total network. If there are two or more components in a parallel branch in a complex network, all of the characteristics of a parallel network apply to this part of the complex network. If there are two or more components in series in the complex network, all of the characteristics of a series network apply to this part of the circuit. The best way to solve a series-parallel network problem is to break it down into separate parallel and series networks. When this separation is done, each of the sections can be simplified in turn. In example 6-1 we will show a simple series-parallel network problem along with each step of the simplification process. Example 6-1 Problem: A 12-ohm resistor is placed in series with a parallel network of 10 ohms and 40 ohms. A 100 volt supply is connected to this series-parallel combination. What is the current through each resistor? What is the total current? Circuit for example 6-1 Solution: Step 1: Isolate R 2 and R 3. Find the equivalent resistance: R eq = (10) (40) 400 = = 8 ohms Ver

26 AM Basic Electronics DC Circuits Analysis Now the circuit is reduced to: Circuit equivalent for example 6-1 Step 2: Find the series equivalent resistance. R t = = 20 ohms 6-2 Ver. 1.0